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Which of the series, and which diverge? Use any method, and give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{1-n}{n 2^{n}}$$

Short Answer

Expert verified
This series diverges.

Step by step solution

01

Write down the problem

We are given the series \( \sum_{n=1}^{\infty} \frac{1-n}{n 2^{n}} \). The task is to determine whether this series converges or diverges.
02

Apply the Ratio Test

The Ratio Test is a common way to determine the convergence of a series. For a series \( \sum a_n \), consider \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If this limit is less than 1, the series converges. If the limit is greater than 1, or infinite, the series diverges. For this series, let \[ a_n = \frac{1-n}{n 2^{n}}. \] Then, \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{1-(n+1)}{(n+1) 2^{n+1}}}{\frac{1-n}{n 2^n}} \right|. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Convergence
Understanding series convergence is crucial when you deal with infinite series. Convergence means that as you add more and more terms to the series, the sum approaches a specific number. This number is called the sum of the series. An convergent series does not just keep growing larger as you add more terms, which is what would happen in a divergent series.

When seeing if a series converges, several tests are commonly used, including the Ratio Test, which was applied in the original exercise. This test involves taking the limit of the absolute value of the ratio of consecutive terms, known as \( \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \). This helps to predict if the series converges without actually summing all the terms, which is useful because infinite series have infinite terms.

In summary, determining a series's convergence tells you whether the series adds up to a finite value or not.
Mathematical Proof
A mathematical proof is a step-by-step logical argument that shows why a statement is true. In the context of series convergence, it establishes whether the proposed series truly converges.

Proof involves:
  • Writing down the problem clearly to understand what you are working with.
  • Using relevant mathematical tests and theorems, such as the Ratio Test used here. This test requires you to compute a limit.
  • Interpreting the result. If the limit is less than 1, the proof concludes that the series converges; if more than 1, it diverges.
Proofs are important because they validate initial conjectures or assumptions using rigorous logic and previously established results, making the conclusions reliable.
Infinite Series
Infinite series are sums that have an infinite number of terms. They are written as \( \sum_{n=1}^{\infty} a_n \), representing the addition of all terms \( a_n \) from \( n=1 \) to infinity.

An infinite series can converge, meaning it settles towards a finite sum, or it can diverge, which means it doesn’t approach any limit. Infinite series can be tricky, as adding an infinite amount of something often suggests it could grow endlessly. However, many series, especially those with terms that shrink in size, can indeed converge.

Working with infinite series often involves applying specific tests to understand their behavior, as manual calculation isn’t feasible. The logical methods applied, such as the Ratio Test, help manage this complexity and predict whether a series converges.

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Most popular questions from this chapter

The first term of a sequence is \(x_{1}=1 .\) Each succeeding term is the sum of all those that come before it: \(x_{n+1}=x_{1}+x_{2}+\cdots+x_{n}\). Write out enough early terms of the sequence to deduce a general formula for \(x_{n}\) that holds for \(n \geq 2\).

Show that if \(M\) is a number less than \(1,\) then the terms of \(\\{n /(n+1)\\}\) eventually exceed \(M .\) That is, if \(M<1\) there is an integer \(N\) such that \(n /(n+1)>M\) whenever \(n>N .\) since \(n /(n+1)<1\) for every \(n,\) this proves that 1 is a least upper bound for \(\\{n /(n+1)\\}\).

Which of the sequences converge, and which diverge? Give reasons for your answers. \(a_{n}=\frac{1+\sqrt{2 n}}{\sqrt{n}}\)

Assume that each sequence converges and find its limit. \(a_{1}=-1, \quad a_{n+1}=\frac{a_{n}+6}{a_{n}+2}\)

To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1

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